An electric toaster is rated 1200 watts at 120 volts. what is the total electrical energy used to operate the toaster for 30 sec
1 answer:
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Answer:
binding energy is 99771 J/mol
Exlanation:
given data
threshold frequency = 2.50 × Hz
solution
we get here binding energy using threshold frequency of the metal that is express as
..................1
here E is the energy of electron per atom and h is plank constant i.e.
and x is binding energy
and here N is the Avogadro constant =
so E will
E =
so put value in equation 1 we get
= 2.50 ×
×
solve it we get
x = 99770.99
so binding energy is 99771 J/mol
Answer:
1. False 2) greater than. 3) less than 4) less than
Explanation:
1)
- As the collision is perfectly elastic, kinetic energy must be conserved.
- The expression for the final velocity of the mass m₁, for a perfectly elastic collision, is as follows:
- As it can be seen, as m₁ ≠ m₂, v₁f ≠ 0.
2)
- As total momentum must be conserved, we can see that as m₂ > m₁, from the equation above the final momentum of m₁ has an opposite sign to the initial one, so the momentum of m₂ must be greater than the initial momentum of m₁, to keep both sides of the equation balanced.
3)
- The maximum energy stored in the in the spring is given by the following expression:
- where A = maximum compression of the spring.
- This energy is always the sum of the elastic potential energy and the kinetic energy of the mass (in absence of friction).
- When the spring is in a relaxed state, the speed of the mass is maximum, so, its kinetic energy is maximum too.
- Just prior to compress the spring, this kinetic energy is the kinetic energy of m₂, immediately after the collision.
- As total kinetic energy must be conserved, the following condition must be met:
- So, it is clear that KE₂f < KE₁₀
- Therefore, the maximum energy stored in the spring is less than the initial energy in m₁.
4)
- As explained above, if total kinetic energy must be conserved:
- So as kinetic energy is always positive, KEf₂ < KE₁₀.
Answer:
0.24 kgm²
Explanation:
= length of the bat = 81.3 cm = 0.813 m
= mass of the bat = 0.96 kg
= distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m
= Period of oscillation = 1.35 sec
= moment of inertia of the bat
Period of oscillation is given as
= 0.24 kgm²